Question: Solve for $x$, $ \dfrac{x + 3}{5x + 10} = -\dfrac{6}{4x + 8} + \dfrac{10}{x + 2} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5x + 10$ $4x + 8$ and $x + 2$ The common denominator is $20x + 40$ To get $20x + 40$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{x + 3}{5x + 10} \times \dfrac{4}{4} = \dfrac{4x + 12}{20x + 40} $ To get $20x + 40$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{4x + 8} \times \dfrac{5}{5} = -\dfrac{30}{20x + 40} $ To get $20x + 40$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{10}{x + 2} \times \dfrac{20}{20} = \dfrac{200}{20x + 40} $ This give us: $ \dfrac{4x + 12}{20x + 40} = -\dfrac{30}{20x + 40} + \dfrac{200}{20x + 40} $ If we multiply both sides of the equation by $20x + 40$ , we get: $ 4x + 12 = -30 + 200$ $ 4x + 12 = 170$ $ 4x = 158 $ $ x = \dfrac{79}{2}$